Question

How do you solve `\frac { 14p } { 5} - 3= \frac { 3p } { 8} + \frac { 7} { 30}`?

Answer 1

`p=1\frac{1}{3}`

Explanation:

`\frac{14p}{5}-3=\frac{3p}{8}+\frac{7}{30}`

Rearrange the terms so the terms with variables are on one side, and the constants are on the other side:

`\frac{14p}{5}-\frac{3p}{8}=\frac{3}{70}+3`

Find a common denominator on the LHS, which is `40`:

`\frac{14p\cdot color(red)(8)}{5\cdot color(red)(8)}-\frac{3p\cdot color(green)(5)}{8\cdot color(green)(5)}=\frac{7}{30}+3`

`\rightarrow\frac{122p}{40}-\frac{15p}{40}=\frac{7}{30}+3`

Change the `3` on the RHS to have a denominator of `30`:

`\frac{112p}{40}-\frac{15p}{40}=\frac{7}{30}+\frac{90}{30}`

Simplify both sides:

`\frac{color(magenta)(97p)}{color(blue)(40)}=\frac{color(blue)(97)}{color(magenta)(30)}`

Cross multiply:

`color(magenta)(97p)\cdot color(magenta)(30)=color(blue)(40)\cdot color(blue)(97)`

Evaluate both sides:

`3880=2910p`

Isolate for `p`:

`p=1\frac{1}{3}`

Answer 2

`p=4/3`

Explanation:

If an equation has fractions, you can get rid of them immediately by multiplying each term by the LCM of the denominators. (The LCD) so that you can cancel the denominators

In this case it is `120`

`(color(blue)(120xx) 14p)/5 - color(blue)(120xx) 3= (color(blue)(120xx)3p)/8 + (color(blue)(120xx)7)/30`

`(color(blue)(cancel(120)^24xx) 14p)/cancel5 - color(blue)(120)xx 3= (color(blue)(cancel(120)^15xx)3p)/cancel8 + (color(blue)(cancel(120)^4xx)7)/cancel30`

`336p -360 = 45p +28`

`336p-45p = 28+360`

`291p = 388`

`p = 388/291`

`p=4/3`